Question

A uniform disc of radius $R$ and mass $M$ is free to rotate about a fixed horizontal axis perpendicular to its plane and passing through its centre. A string is wrapped over its rim and a block of mass $m$ is attached to the free end of the string. The block is released from rest. If string does not slip on the rim then find the acceleration of the block. Neglect the mass of the string.

Answer 1

Since string does not slip in the tangential acceleration of the point on the rim which is in contact with the string is equal to the acceleration of the block. Let angular acceleration of the disc about the axis be $\alpha$, hence acceleration of the block,$a=\alpha R$
$\Rightarrow mg-T=m\alpha R$ as $a=\alpha R$ ...$\left(1\right)$
Torque on the disc of $\tau ={\tau }_{Tension}+{\tau }_{Mg}$
$\Rightarrow I\alpha =TR$ as ${\tau }_{mg}=0$
Where $I=M.I.$ of the disc about the axis.
$\Rightarrow T=I\alpha$ ...$\left(2\right)$
Eliminating $T$ from $\left(1\right)$ and $\left(2\right)$,
$\alpha =\frac{mg}{(mR+\frac{I}{R})}=\frac{2mg}{(2m+M)R}$ As $I=\frac{m{R}^2}{2}$ Hence $a=\frac{2mg}{2m+M}$